Question: Let $g(x)=\sqrt{x^3+8x}$, for $x>0$. Where does $g$ have critical points? Choose all answers that apply: Choose all answers that apply: (Choice A) A $x=\dfrac{3}{8}$ (Choice B) B $x=\dfrac{8}{3}$ (Choice C) C $x=\sqrt{8}$ (Choice D) D $g$ has no critical points.
Solution: A critical point of $g$ is a point in the domain of $g$ where the derivative is either equal to zero or undefined. So in order to find the critical points of $g$, let's find its derivative. $\begin{aligned} g'(x)&=\dfrac{d}{dx}\left[ \sqrt{x^3+8x} \right] \\\\ &=\dfrac{1}{2\sqrt{x^3+8x}} \cdot \dfrac{d}{dx}[x^3+8x] \\\\ &=\dfrac{3x^2+8}{2\sqrt{x^3+8x}} \end{aligned}$ Now let's look for $x$ -values where $g'$ is zero or undefined. $\dfrac{3x^2+8}{2\sqrt{x^3+8x}}=0$ has no solution, so $g'$ is never equal to $0$. $\dfrac{3x^2+8}{2\sqrt{x^3+8x}}$ is undefined at $x=0$, but this is outside of our given domain. In conclusion, $g$ has no critical points.